This is the final preprint version of a paper which appeared in Algebra Universalis, 68 (2012) 17-37. The published version is accessible to subscribers at http://dx.doi.org/10.1007/s00012-012-0191-2 .

Isotone maps on lattices

G. M. Bergman and G. Gratzer¨

Abstract. Let L = (Li | i ∈ I) be a family of lattices in a nontrivial lattice vari- ety V, and let ϕi : Li → M, for i ∈ I, be isotone maps (not assumed to be lattice homomorphisms) to a common lattice M (not assumed to lie in V). We show that the maps ϕi can be extended to an isotone map ϕ: L → M, where L = FreeV L is the free product of the Li in V. This was known for L = V, the variety of all lattices. The above free product L can be viewed as the free lattice in V on the partial lattice P formed by the disjoint union of the Li. The analog of the above result does not, however, hold for the free lattice L on an arbitrary partial lattice P . We show that the only codomain lattices M for which that more general statement holds are the complete lattices. On the other hand, we prove the analog of our main result for a class of partial lattices P that are not-quite-disjoint unions of lattices. We also obtain some results similar to our main one, but with the relationship lat- tices : orders replaced either by semilattices : orders or by lattices : semilattices. Some open questions are noted.

1. Introduction

By Yu. I. Sorkin [15, Theorem 3], if L = (Li | i ∈ I) is a family of lattices and ϕi : Li → M are isotone maps of the lattices Li into a lattice M, then there exists an isotone map ϕ from the free product Free L of the Li to M that extends all the ϕi. (There are some difficulties with Sorkin’s original proof; but a simple, correct proof is given by G. Gr¨atzer,H. Lakser and C. R. Platt in [11, §4].)

2010 Mathematics Subject Classification: Primary: 06B25. Secondary: 06B20, 06B23. Key words and phrases: Free product of lattices; varieties, prevarieties and quasivarieties of lattices; isotone map; free lattice on a partial lattice; semilattice. 2 G. M. Bergman and G. Gr¨atzer Algebra univers.

Our main result, proved in Section 2, is a generalization of this fact, with Free L replaced by FreeV L, the free product of the Li in any nontrivial va- riety V of lattices containing them — though not necessarily containing M. (In Section 3, we explore some variants of our proof of this result.) We may regard FreeV L as the free lattice in V on the partial lattice P given by the disjoint union of the Li. Does the analog of the above result hold for more general partial lattices P and their free lattices L? In Section 4 we find that the lattices M such that this statement holds for all partial lattices P are the complete lattices. On the other hand, we describe in Section 5 a class of partial lattices P , related to but distinct from the class considered in Section 2, for which the full analog of the result of that section holds. Since semilattices lie between orders and lattices, it is plausible that state- ments similar to our main result should hold, either with “lattice” weakened to “semilattice”, or with “lattice” unchanged but “isotone map” strengthened to “semilattice homomorphism”. In Section 6 we shall find that the former statement is easy to prove. In that section and Section 7, we obtain sev- eral approximations to the latter statement; we do not know whether the full statement holds. The reader familiar with the concepts of quasivariety and prevariety will find that the proofs given in this note for varieties of lattices in fact work for those more general classes. However, varieties are not sufficient for the result of Section 7, so we develop that in terms of prevarieties. In Section 8 we note some open questions. For general definitions and results in lattice theory, see [8] or [9]. For another context in which isotone maps among lattices have appeared in the literature—in this case, isotone sections of lattice homomorphisms—see R. Freese, J. Jeˇzek,and J. B. Nation [6, pp. 99–107, 133] and R. Freese [5]. We are indebted to R. Freese for pointing us to those works, and for the related observation following Corollary 4 below; and to A. V. Kravchenko for an extensive correspondence regarding Sorkin’s original argument.

2. Extending isotone maps to free product lattices

Let V be a nontrivial lattice variety, that is, a variety V of lattices having a member with more than one element. Let L = (Li | i ∈ I) be a fam- ily of lattices in V, and L = FreeV L their free product in V. Finally, let (ϕi : Li → M | i ∈ I) be a family of isotone maps into a lattice M, not as- sumed to lie in V. To show that the ϕi have a common extension to L, it suffices, by the 0 universal property of L, to find some L ∈ V such that each map ϕi factors 0 Li → L → M, where the first map is a lattice homomorphism, and the second an isotone map not depending on i. So let us, for now, forget free products, and obtain such a lattice L0. Vol. 00, XX Isotone maps on lattices 3

We first note (as remarked in [11, last paragraph] for V = L) that this is easy if M has a least element, or more generally, if its subsets ϕi(Li) have a common lower bound e ∈ M. In that case, we begin by enlarging all the Li to lattices L¯i = {ei} + Li, where ei is a new least element, and extend the ϕi to maps 0 Q ϕ¯i : L¯i → M mapping ei to e. Now let L be the sublattice of (L¯i | i ∈ I) consisting of the elements x = (xi | i ∈ I) such that xi = ei for all but 0 finitely many i; and let us map each Li into L by the homomorphism sending x ∈ Li to the element having i-component x, and j-component ej for all j 6= i. We now map L0 to M using the isotone map ψ given by W 0 ψ(x) = (ϕ ¯i(xi) | i ∈ I ) for x = (xi | i ∈ I) ∈ L . (1) This infinite join is defined because all but finitely many of the joinands are 0 e; and it is easy to verify that for each i, the composite map Li → L → M is the given isotone map ϕi, as required. If, rather, the ϕi(Li) have a common upper bound, the dual construction is, of course, available. In the absence of either sort of bound, we shall follow the same pattern of adjoining to the Li elements ei with a common image e in M (this time an arbitrary element of that lattice); but that construction takes a bit more work, as does the one analogous to the definition (1) of the isotone map ψ : L0 → M. The first of these steps is carried out in the following lemma (where L corre- sponds to the above Li). Lemma 1. Let ϕ: L → M be any isotone map of lattices, and e any ele- ment of M. Then there exists a lattice extension L¯ of L, and an isotone map ϕ¯: L¯ → M extending ϕ, such that e ∈ ϕ¯(L¯). Moreover, L¯ can be taken to lie in any nontrivial lattice variety V containing L.

Proof. Let L¯ = L × C2 × C2, where C2 is the 2-element lattice {0, 1}, and embed L in L¯ by x 7→ (x, 0, 1). Defineϕ ¯: L¯ → M by   ϕ¯(x, 0, 1) = ϕ(x),    ϕ¯(x, 1, 0) = e, (2)  ϕ¯(x, 0, 0) = ϕ(x) ∧ e,    ϕ¯(x, 1, 1) = ϕ(x) ∨ e.

It is easy to check thatϕ ¯ is isotone, and it clearly has e in its range. Since C2 = {0, 1} belongs to every nontrivial variety of lattices, L¯ will belong to any nontrivial variety V containing L.  The next lemma gives the construction we will use to weld our I-tuple of isotone maps into one map.

Lemma 2. Let M be a lattice, e any element of M, and I a nonempty set. Let M 0 be the sublattice of M I consisting of those elements f such that f(i) = e 4 G. M. Bergman and G. Gr¨atzer Algebra univers. for all but finitely many i ∈ I. Then there exists a map ψ : M 0 → M such that ψ is isotone (3) and For every i ∈ I, and every f ∈ M 0 satisfying f(j) = e for all j 6= i, (4) we have ψ(f) = f(i). Proof. For f ∈ M 0, define  V( f(i) | i ∈ I ) if f(i) ≤ e for all i ∈ I, ψ(f) = (5) W  ( f(i) | i ∈ I, f(i)  e ) otherwise. These meets and joins are defined because for each f ∈ M 0, there are only finitely many distinct values f(i). It is easy to see that ψ satisfies (4). To obtain (3), observe that For f ≤ g in M 0, we have {i | f(i) 6≤ e} ⊆ {i | g(i) 6≤ e}. (6) Hence given f ≤ g, there are three possibilities: Either the definitions of ψ(f) and ψ(g) both fall under the first case of (5), or they both fall under the second, or that of ψ(f) falls under the first and that of ψ(g) under the second. If both fall under the first case, then ψ(f) ≤ ψ(g) because the meet opera- tion of M is isotone. If both fall under the second, the same conclusion follows using the fact that the join operation is isotone, together with the fact that bringing in more joinands, as can happen in view of (6), yields a join greater than or equal to what we would get without those additional terms. Finally, if the evaluation of ψ(f) falls under the first case and that of ψ(g) under the second, we may choose, in view of the latter fact, an i such that g(i) 6≤ e. Then ψ(f) ≤ f(i) ≤ g(i) ≤ ψ(g), (7) completing the proof of (3).  We can now fill in the proof of our main theorem.

Theorem 3. Let V be a nontrivial variety of lattices, L = (Li | i ∈ I) a family of lattices in V, and (ϕi : Li → M | i ∈ I) a family of isotone maps from the Li to a lattice M not necessarily in V. Then there exists an isotone map ϕ: FreeV L → M whose restriction to each Li ⊆ FreeV L is ϕi.

Proof. Choose any element e ∈ M, and extend each ϕi as in Lemma 1 to a mapϕ ¯i : L¯i → M on a lattice extension L¯i ⊇ Li in V, so that some ei ∈ L¯i is Q mapped byϕ ¯i to e ∈ M. Now map each Li into (L¯i | i ∈ I) by sending every element x ∈ Li to the element having i-th coordinate x, and j-th coordinate ej for all j 6= i. These maps are lattice homomorphisms, hence together they Q induce a homomorphism FreeV L → (L¯i | i ∈ I). Moreover, this map has Vol. 00, XX Isotone maps on lattices 5

0 range in the sublattice L of elements whose j-coordinates are ej for almost all j, since the image of each Li lies in that sublattice. Q I Q Mapping L¯i to M by the isotone map ϕ¯i, we see that the above 0 Q 0 I sublattice L ⊆ L¯i is carried into the sublattice M ⊆ M of Lemma 2. Bringing in the isotone map f : M 0 → M of that lemma, we get our desired 0 0 isotone map ϕ as the composite FreeV L → L → M → M. It follows from (4) that the restriction of ϕ to each Li is ϕi.  We note a curious consequence of the fact that the M of Theorem 3 need not lie in V.

Corollary 4. Let L = (Li | i ∈ I) be a family of lattices in a nontrivial lattice variety V, let FreeV L be their free product in V, and let Free L be their free product in the variety L of all lattices. Then there exists an isotone map FreeV L → Free L which acts as the identity on each Li. In particular, for any nontrivial lattice variety V and any set X, there exists an isotone map FreeV(X) → Free(X)(where these denote the free lattices on the set X in V and in L respectively), which acts as the identity map on X.

Proof. For the first statement, apply Theorem 3 to the inclusions of the Li in Free L. The second is the case of the first where all Li are one-element lattices.  The existence of a map as in the last sentence of the above corollary also follows from results in the literature, which in fact show that it can be taken to be a set-theoretic right inverse to the natural lattice homomorphism Free(X) → FreeV(X). Namely, it is shown by R. Freese [4] that free lattices in any lattice variety satisfy an elegant order-theoretic property called “finite separability”, and it is shown in R. Freese and J. B. Nation [5] (where finite separability was called “condition (4)”; for a rewrite of that development, us- ing the present name, see R. Freese, J. Jeˇzek,and J. B. Nation [6, Lemma 5.6]) that this property allows one to construct isotone right inverses to surjective homomorphisms. It would be interesting to know whether the isotone map of the first para- graph of the above corollary can similarly be taken to be a right inverse to the natural homomorphism Free L → FreeV L. The lattice FreeV L will not be finitely separable if the Li are not; but perhaps it has some corresponding relative property.

3. Digression: sketches of some alternate proofs of Theorem 3

The definition (5) of the isotone map ψ used in the proof of Lemma 2 is clearly asymmetric in the meet and join operations. We sketch below a variant proof of Theorem 3 which uses a function that is symmetric in these operations — but lacks instead (when |I| > 2) the symmetry in the family of lattices Li which the proof given above clearly has. 6 G. M. Bergman and G. Gr¨atzer Algebra univers.

We shall then show that one cannot have it both ways: a map of the required sort having both sorts of symmetry does not, in general, exist. However, we show that we can get such a map if M lies in the given variety V. This section will be sketchier than the rest of the paper. In particular, we will be informal about our two sorts of symmetry; though in the next-to-last paragraph, we will indicate how to make these considerations precise. Our new proof of Theorem 3 starts with a generalization of the construction of Lemma 1. Namely, suppose we are given isotone maps of two lattices into a common lattice, ϕi : Li → M for i = 0, 1. Let

0 L = L0 × L1 × C2 × C2. (8)

0 Then taking any e0 ∈ L0, e1 ∈ L1, we can embed our two lattices in L by the homomorphisms 0 L0 → L acting by x 7→ (x, e1, 1, 0), (9) 0 L1 → L acting by y 7→ (e0, y, 0, 1). Now define the isotone map ϕ0 : L0 → M by

 0  ϕ (x, y, 1, 0) = ϕ0(x),   ϕ0(x, y, 0, 1) = ϕ (y), 1 (10) ϕ0(x, y, 0, 0) = ϕ (x) ∧ ϕ (y),  0 1  0  ϕ (x, y, 1, 1) = ϕ0(x) ∨ ϕ1(y).

0 Clearly, ϕ acts on the embedded images of the Li by the ϕi; and as before, 0 since C2 belongs to every nontrivial variety of lattices, L belongs to any non- trivial variety V containing the Li. This, in fact, gives us Theorem 3 for |I| = 2 by a construction symmetric both in meet and join, and in our family of lattices. Now suppose more generally that we have lattices Li ∈ V and isotone maps ϕi : Li → M indexed by an arbitrary set I. Assuming without loss of generality that I is an ordinal, we shall construct the desired L0 ∈ V and isotone map ϕ0 : L0 → M by a recursive transfinite iteration of the above construction. It is the recursion that will lose us our symmetry in the Li, via the arbitrary choice of an identification of I with an ordinal, i.e., of a well-ordering on I. To describe the recursion, let 1 < k ≤ I, and assume that we have con- 0 structed lattices L(j) for all 1 ≤ j < k, which satisfy 0 0 0 L0 = L(1) ⊆ L(2) ⊆ · · · ⊆ L(j) ⊆ · · · , (11)

0 together with lattice embeddings Li → L(j) for i < j, and isotone maps 0 L(j) → M, and that these form a coherent system, in the sense that for 0 0 0 0 i < j < j , the composite Li → L(j) ⊆ L(j0) is the embedding Li → L(j0), and 0 0 0 the composite L(j) ⊆ L(j0) → M is the isotone map L(j) → M; and, finally, 0 such that for every i < j, the composite Li → L(j) → M is the given isotone map ϕi. Vol. 00, XX Isotone maps on lattices 7

If k is a successor ordinal, k = j + 1, we apply the |I| = 2 case of our 0 construction, described in (8)-(10), to the pair of lattices L(j) and Lj and their isotone maps to M, calling the resulting lattice

0 0 L(j+1) = L(j) × Lj × C2 × C2, (12)

0 and identifying L(j) with its image therein under the first map of (9). If, on 0 0 the other hand, k is a limit ordinal, we let L(k) be the union of the L(j) over all j < k. In each case, the asserted properties are immediate. Thus, we can 0 carry our construction up to k = I, the resulting lattice L(I) being our desired L0. What are the consequences of the different kinds of symmetry of the con- struction of the preceding section and the one just sketched? Because the former was symmetric in the Li, we can deduce, for instance, that in the final statement of Corollary 4, if X is finite, then the isotone map FreeV(X) → Free(X) can be taken to respect the actions of the symmetric group Sym(X) on these two lattices. (Why assume X to be finite? So that in applying Lemma 1, we can choose an e ∈ M = Free(X) invariant under that group action, say the join of the given generators. Alternatively, without this finiteness assumption, if we choose any x0 ∈ X and perform our construction with e = x0, we can get our map to respect the action of Sym(X − {x0}).) On the other hand, using our new construction we can deduce that if V is closed under taking dual lattices, we can, instead, in that same final statement of Corollary 4, take the isotone map FreeV(X) → Free(X) to respect the anti- automorphisms of the domain and codomain that fix the free generators but interchange meet and join. (Again, we have to decide what to use for our distinguished elements e0, e1 at each application of (9). In this case, we may, at each such step, take e0 to be any of the preceding generators, while for e1 we have no choice but to use the generator we are adjoining.) Let us now show that for |I| = 3, we cannot get a construction with both sorts of symmetry. If we could, then letting D denote the variety of distributive lattices, we could get an isotone map ϕ: FreeD(3) → Free(3) respecting all permutations of the generators, and also respecting the anti-automorphisms that interchange meets and joins. Now FreeD(3) has an element invariant under all these symmetries; namely, writing its three generators a, b, c, the element

(a ∧ b) ∨ (b ∧ c) ∨ (c ∧ a) = (a ∨ b) ∧ (b ∨ c) ∧ (c ∨ a). (13)

On the other hand, for any set X, the only elements of Free(X) that can be invariant under an anti-automorphism are the given free generators; this follows from the fact that every element of Free(X) other than those generators is either meet-reducible or join-reducible, but never both. (Cf. [9, condition (W) on p. 477, and Corollary 534(iii)].) So Free(3) has no element with both sorts of symmetry to which one could send the element given by (13). 8 G. M. Bergman and G. Gr¨atzer Algebra univers.

Finally, let us show that we can get both sorts of symmetry if the lattice M lies in V. (Of course, this restriction makes it impossible to use the result to prove a version of Corollary 4.) We record in the next lemma the raw construction used in the proof. Though that lemma requires M to lie in V, it does not require the same of the Li. But it is easy to see that if we add 0 the assumption that the Li lie in V, the lattice L obtained will lie there as well, hence the construction will induce, as in Theorem 3, an isotone map ϕ: FreeV L → M acting as ϕi on each Li. Moreover, the construction clearly has all the asserted symmetries. (We remark that the factor FreeV(I) in the construction reduces, when |I| = 2, to the lattice C2 × C2 of (8). So one could say it was the fact that all nontrivial lattice varieties have the same 2-generator free lattice that allowed us to get the doubly symmetric construction in that two-lattice case with no added restriction on M.)

Lemma 5. Let L = (Li | i ∈ I) be a family of lattices, let M be a lattice, and for each i ∈ I, let ϕi : Li → M be an isotone map. Let V be a lattice variety containing M, and in the free lattice FreeV(I), let the i-th generator be denoted gi for each i. Let

0 Q L = (Li | i ∈ I) × FreeV(I). (14)

0 Suppose we choose, for each i ∈ I, a lattice homomorphism ξi : Li → L which takes every x ∈ Li to an element whose i-th coordinate is x and whose coordinate in FreeV(I) is the generator gi. (For instance, such a family of homomorphisms ξi can be determined by fixing an element ej in each Lj, and letting ξi(x) have, in addition to the two coordinates just specified, j-th coordinate ej for all j ∈ I − {i}.) Finally, let ϕ0 : L0 → M be the function taking each pair (x, w), where Q x = (xi | i ∈ I) ∈ (Li | i ∈ I) and w ∈ FreeV(I), (15)

I to w¯(ϕi(xi) | i ∈ I), where w¯ : M → M is the operation of evaluating the lattice expression w ∈ FreeV(I) at I-tuples of elements of M. 0 0 Then ϕ is an isotone map, and for each i ∈ I, ϕ ξi = ϕi.

Sketch of proof. The final equation is clear. To show that ϕ0 is isotone, let (x, w) ≤ (x0, w0) in L0. Then we claim that ϕ0(x, w) ≤ ϕ0(x0, w) ≤ ϕ0(x0, w0); 0 0 0 in other words, thatw ¯(ϕi(xi)) ≤ w¯(ϕi(xi)) ≤ w¯ (ϕi(xi)). The first inequality holds becausew ¯, a derived lattice operation, is isotone; the second, because 0 evaluation at (ϕi(xi)), a lattice homomorphism FreeV(I) → M, is isotone. 

We remark that the isotone map ϕ0 of the above lemma is not, in general, a lattice homomorphism. (For instance, let I = {0, 1}, let L0 = L1 = M = C2, let the ϕi : Li → M be the identity map, and let V be any nontrivial lattice variety. Denoting the generators of FreeV(I) by g0 and g1, we note that

0 ((0, 1), g1 ∧ g2) ∨ ((1, 0), g1 ∧ g2) = ((1, 1), g1 ∧ g2) in L . (16) Vol. 00, XX Isotone maps on lattices 9

The map ϕ0 takes each of the joinands on the left to 0, but the term on the right to 1.) We have discussed symmetries of our construction informally above, leav- ing it to the reader to see that a construction with a certain sort of symmetry would imply corresponding properties of the maps constructed. These con- siderations can, of course, be formalized. If we describe our constructions as functors on appropriate categories of systems of lattices, isotone maps, and distinguished elements, then constructions with various sorts of symmetry al- low us to strengthen the conclusion of Theorem 3 to say that we have functors respecting certain additional structure on those categories. We shall not go into these details here, however. We turn now to some variants of our main result.

4. When does the same result hold for the inclusion of a general partial lattice P in its free lattice L?

If the lattice M of Theorem 3 happens to be a complete lattice, the con- clusion of that theorem follows from a much more general fact: Any isotone map from an order P into a complete lattice can be extended to any order extension Q of P . In other words, in the category of orders, complete lattices are injective with respect to inclusions of orders. The inclusions of orders are not, up to isomorphism, the only monomor- phisms in the category of orders and isotone maps. B. Banaschewski and G. Bruns [1] characterize the inclusions category-theoretically among the mono- morphisms, calling them the strict monomorphisms, and they formulate the above result as the statement that every complete lattice (in their terminology, every complete partially ordered set) is a “strict injective”; to which they also prove the converse [1, Proposition 1, (i)⇐⇒(ii)]. Theorem 3 can thus be looked at as saying that if P is the disjoint union of a family of lattices Li belonging to a variety V, regarded as a partial lattice, then the inclusion of P in its free lattice L = FreeV P behaves a little better than a general inclusion of orders, in that isotone maps of P to arbitrary lattices, and not only to complete lattices, can be extended to L. In contrast, we shall see below that the inclusion of a general partial lattice P in its free lattice Free P behaves no better in this way than do arbitrary extensions of orders, at least insofar as isotone maps to lattices are concerned. (For the concepts of a partial lattice and of the free lattice on such an object, see [8, Section I.5], [9, Sections I.5.4-I.5.5].) We begin with the building blocks from which the “test cases” showing this will be put together.

Lemma 6. Let B be a boolean lattice with > 2 elements. Then as an extension of B − {0, 1}, the lattice Free(B − {0, 1}) is isomorphic to B. 10 G. M. Bergman and G. Gr¨atzer Algebra univers.

Proof. Let P = B − {0, 1}. The only joins that P is missing are those that in B yield 1; likewise, the only missing meets are those that yield 0. We shall show that all pairs of elements which had join 1 in B give equal joins in any lattice L into which we map P by a homomorphism of partial lattices. By symmetry, the dual statement holds for 0 and meets. Hence the free lattice on P just restores these two elements, i.e., it is naturally isomorphic to B. So suppose we are given a map of P into a lattice L, which preserves the meets and joins of P . By abuse of notation, we shall use the same symbols for elements of P and their images in L. Let us first consider any two elements a, b ∈ P which are distinct in P from each other and from each other’s complements, and compare the joins a ∨ ac and b ∨ bc in L (writing ( )c for complementation in B). Note that in B, we have

a = (a ∧ b) ∨ (a ∧ bc) and ac = (ac ∧ b) ∨ (ac ∧ bc). (17)

If the four meets appearing in these two expressions are all nonzero, then they belong to P , and the relations (17) hold there, and hence in L. In this situation, if we expand a ∨ ac in L using these two formulas, we can rearrange the result as ((a∧b)∨(ac ∧b))∨((a∧bc)∨(ac ∧bc)), which (by (17) with a and b interchanged) simplifies to b ∨ bc, giving the desired equality. On the other hand, if any of the four pairwise meets of a and ac with b and bc is zero, this can, under the assumptions made in the preceding paragraph, be true only of one such meet; say a ∧ b = 0. Then we can repeat the above computation, everywhere omitting “(a∧b)∨”. (Thus, we have a version of (17) with the first equation simplified to a = a ∧ bc, and the second unchanged.) With this slight modification, our computation still works, and again gives a ∨ ac = b ∨ bc. So let us write i for the common value, for all a ∈ P , of a ∨ ac ∈ L. We now consider two elements a, b ∈ P which are not assumed to be complements, but whose join in B is 1. This relation implies that b ≥ ac; note that both these terms lie in P . Hence in L we have a ∨ b ≥ a ∨ ac = i, while the reverse inequality holds because a ≤ i, b ≤ i. Thus, a ∨ b = i, completing the proof that all pairs of elements having join 1 in B have the same join, namely i, in L. 

Let us now consider, independent of the above result, the same inclusion B − {0, 1} ⊆ B in the context of isotone maps.

Lemma 7. Let M be a lattice, X a nonempty subset of M, and B the free Boolean lattice on X. Then there exists an isotone map ϕ: B − {0, 1} → M with the property that

The pairs of elements y, z ∈ M such that ϕ can be extended to an isotone map ϕ¯: B → M taking 0 to y and 1 to z, are precisely those (18) for which y is a lower bound, and z an upper bound, for X in M. Vol. 00, XX Isotone maps on lattices 11

Proof. Let B have free generators gx for x ∈ X. For every a ∈ B − {0, 1}, let ϕ(a) be the join in M of all elements of the form

x1 ∧ · · · ∧ xn (19) where n ≥ 1, and x1, . . . , xn are distinct elements of X such that for some choice of ε1, . . . , εn ∈ {1, c}, we have a ≥ gε1 ∧ · · · ∧ gεn . x1 xn (20) Here for b ∈ B, bε denotes b if ε = 1, and bc if ε = c. Because a 6= 0, the set of instances of (20) is nonempty, hence so is the set of joinands (19). This set is in general infinite; however, if we take the least subset X0 ⊆ X such that a is in the Boolean sublattice generated by the gx with x ∈ X0, then X0 is finite and (since a 6= 0, 1) nonempty; and we find that the irredundant relations (20) (those relations (20) from which ε no meetand can be dropped) involve only terms gx with x ∈ X0. Thus, each expression (19) in our description of ϕ(a) is majorized by one that arises from one of these finitely many irredundant relations (20); so the join describing ϕ(a) is effectively a finite join, and so exists in M. It is not hard to see from our definition that ϕ is isotone, and that for all c x ∈ X, ϕ(gx) = ϕ(gx) = x. Suppose now that we have an extensionϕ ¯: B → M of this isotone map ϕ. Then for every x ∈ X,ϕ ¯(1) ≥ ϕ(gx) = x, soϕ ¯(1) is an upper bound of X. Conversely, any upper bound for X in M will majorize all elements (19), and hence all joins of such elements, hence will indeed be an acceptable choice for a value ofϕ ¯(1) makingϕ ¯ isotone. Though our construction of ϕ is not symmetric in ∨ and ∧, the duals of these observations are easily seen to hold, so the choices forϕ ¯(0) are, likewise, the lower bounds of X.  Note that (18) above can be summarized as saying that The upper and lower bounds in M of ϕ(B − {0, 1}) are the same as (21) the upper and lower bounds in M of X. In our next result, for any two partial lattices P and Q, we will denote by P + Q the disjoint union of P and Q, made a partial lattice using the partial meet and join operations of P and Q, together with the further meet and join relations corresponding to the condition that every element of P be majorized by every element of Q (namely, p ∧ q = p and p ∨ q = q for all p ∈ P, q ∈ Q). It is not hard to see that Free(P + Q) =∼ Free P + Free Q. (22) Theorem 8. Let M be a lattice. Then the following conditions are equivalent.

M is complete. (23) Any isotone map from a partial lattice P to M can be extended to an (24) isotone map Free P → M. 12 G. M. Bergman and G. Gr¨atzer Algebra univers.

For B a free Boolean lattice on a nonempty set, any isotone map B1 − {0, 1} → M can be extended to an isotone map B → M; and for B1, B2 any two free Boolean lattices on nonempty sets, any isotone (25) map (B1 − {0, 1}) + (B2 − {0, 1}) → M can be extended to an isotone map B1 + B2 → M. Proof. (23)=⇒(24) is a case of [1, Proposition 1, (i)=⇒(ii)], which says that every complete lattice is injective with respect to inclusions of orders. In view of Lemma 6 and (22), the implication (24)=⇒(25) is clear. To complete the argument, assume (25). Calling on the first statement of (25), together with the case X = M of the preceding lemma, we see that M must have a greatest and a least element. Now take any nonempty subset X1 ⊆ M, let X2 be the set of its upper bounds (which is nonempty, since M has a greatest element), let B1 be the free Boolean lattice on X1, and let B2 be the free Boolean lattice on X2. Map B1−{0, 1} and B2−{0, 1} into M by maps ϕ1, ϕ2 satisfying (18) with respect to X1 and X2, respectively. By the equivalence of (18) and (21), ϕ1(B1 − {0, 1}) is majorized by all upper bounds of X1, i.e., by all elements of X2, hence (again using that equivalence) by all elements of ϕ2(B2 − {0, 1}); so ϕ1 and ϕ2 together constitute an isotone map ϕ:(B1 − {0, 1}) + (B2 − {0, 1}) → M. Extending this to the free lattice B1 + B2 on that partial lattice, we see that the image of the 1 of B1 (and likewise that of the 0 of B2) will be both an upper bound of X1 and a lower bound of X2, hence must be a least upper bound of X1. So M is upper semicomplete. By symmetry (or by the known fact that in a lattice with 0 and 1, upper semicompleteness and lower semicompleteness are equivalent), M is also lower semicomplete, establishing (23). 

5. Lattices amalgamated over convex retracts

The results of the preceding section show that Theorem 3, looked at as a property of the inclusion of a certain kind of partial lattice P in FreeV P , does not go over to the inclusion of a general partial lattice P in its free lattice. Can we describe other interesting partial lattices P for which it does? In proving Theorem 3, after reducing to the case where the given lattices contained elements ei that mapped to the same element of M, we effectively proved that the free lattice on the union of those lattices with amalgamation of the ei had the desired extension property. The next theorem will slightly generalize this result, replacing the singletons {ei} with any family of isomor- phic sublattices that are both retracts of the Li, and convex therein. We will need the following observation.

Lemma 9. Let M be a lattice, and % a lattice-theoretic retraction of M to a convex sublattice. Then if an element x ∈ M is majorized by some element of %(M), then it is majorized by %(x). Vol. 00, XX Isotone maps on lattices 13

Proof. Say x ≤ r ∈ %(M). Applying % to this relation, and taking the join of the original relation with the resulting one, we get x ∨ %(x) ≤ r. Hence x ∨ %(x) lies in the interval between %(x) and r, so as %(M) is assumed convex, x∨%(x) ∈ %(M). This means that x∨%(x) is fixed under the idempotent lattice homomorphism %; but its image under that map is %(x) ∨ %(x) = %(x). Thus, x ∨ %(x) = %(x), which is equivalent to the desired conclusion x ≤ %(x).  In the above lemma, the assumption that % is a lattice homomorphism could have been weakened to say that it is a join-semilattice homomorphism. We have stated it as above for conceptual simplicity, and because in the proof of the next result, the maps %i must be lattice homomorphisms anyway.

Theorem 10. Let (Li | i ∈ I) be a family of lattices which are disjoint except for a common sublattice K, which is convex in each Li, and is a retract of each Li via a lattice-theoretic retraction %i : Li → K. Let P denote the partial lattice given by the union of the Li with amalgama- tion of the common sublattice K, and let L = FreeV P , where V is any variety containing all the Li. Then for any lattice M (not necessarily belonging to V) given with isotone maps ϕi : Li → M agreeing on K, there exists an isotone map ϕ: L → M extending all the ϕi. In other words, every isotone map P → M extends to L.

Proof. Let us assume that I does not contain the symbol 0, and use 0 to index Q 0 the factor K in K × ( Li | i ∈ I). Now let L denote the sublattice of that direct product consisting of those elements f such that f(i) = f(0) for almost 0 all i, and %i(f(i)) = f(0) for all i. Then we can map each Li into L by sending x ∈ Li to the element having i-th coordinate x, and having %i(x) for all other coordinates (including the 0-th coordinate). These maps are lattice homomorphisms (this is where we need the %i to be lattice homomorphisms and not just join-semilattice homomorphisms), which agree on K; hence they extend to a lattice homomorphism L → L0. We shall now map L0 isotonely to M using the idea of Lemma 2. Namely, given f ∈ L0, we define V  ( ϕi(f(i)) | i ∈ I ) if for all i ∈ I, f(i) ≤ f(0), ψ(f) = (26) W  ( ϕi(f(i)) | i ∈ I, f(i) 6≤ f(0) ) otherwise.

These are defined because for each f, all but finitely many i ∈ I have ϕi(f(i)) equal to the common image of f(0) in M under the ϕj. (Recall that f(0) ∈ K, and all ϕj agree on K.) We now claim that ψ is isotone, (27) and for every i ∈ I, and every f ∈ L0 such that f(j) = f(0) for all j 6= i, (28) we have ψ(f) = ϕi(f(i)). 14 G. M. Bergman and G. Gr¨atzer Algebra univers.

Assertion (28) is clear. The proof of (27) is exactly like that of the corre- sponding statement, (3), in the proof of Lemma 2, once we know the analog of (6), namely for f ≤ g in L0, we have {i | f(i) 6≤ f(0)} ⊆ {i | g(i) 6≤ g(0)}. (29) To prove (29), consider any i not lying in the right-hand side. Then f(i) ≤ g(i) ≤ g(0) ∈ %(M), (30) so by Lemma 9, f(i) ≤ %(f(i)) = f(0), showing that i also fails to lie in the left-hand set. Composing ψ with the map L → L0 of the first paragraph of this proof, we get our desired isotone map L → M.  (If we think of the constant e of Lemma 2 as “sea level”, then the f(0) of the above proof brings in “tides”.) We remark that though in the free-lattice-with-amalgamation L of the above proof, K is necessarily a retract, since it was a retract in each of the Li, it does not follow similarly that K is convex in L. To see this, let us first note an example of a lattice L0 having a sublattice K which is convex and a retract in each of two sublattices L0 and L1 containing K, but is not convex in the sublattice that these generate. Let L0 be the lattice of all subspaces of a 3- dimensional vector space V over any field, let K = {{0}, a} where a is a 2-dimensional subspace of V , and let each Li be the sublattice generated by a and a 1-dimensional subspace bi not contained in a, with b0 6= b1. Then the stated hypotheses are satisfied, but 0 < (b0 ∨ b1) ∧ a < a, so K is not convex in the lattice generated by L1 and L2. It easily follows that in the free product L of L0 and L1 with amalgamation of K = {0, 1}, we likewise have 0 < (b0 ∨ b1) ∧ a < a with the middle term not in K.

6. Semilattice variants—two easy results

In our main theorem, free products of a family of lattices Li, whose normal role is to admit a lattice homomorphism extending a given family of lattice homomorphisms on the Li, were made to do the same for isotone maps (homo- morphisms of orders). One might expect it to be easier to get similar results if the gap between lattices and orders is replaced by one of the smaller gaps between lattices and semilattices, or between semilattices and orders. For the latter case, the result is indeed easy; it is only for parallelism with our other results that we dignify it with the title of theorem.

Theorem 11. Let (Li | i ∈ I) be a family of join-semilattices, and ϕi : Li → M a family of isotone maps from the Li to a common join-semilattice. Let L de- note the free product of the Li as join-semilattices. Then there exists an isotone map ϕ: L → M whose restrictions to the Li ⊆ L are the ϕi. Vol. 00, XX Isotone maps on lattices 15

Proof. The general element x ∈ L is a formal join xi1 ∨ · · · ∨ xin of elements xim ∈ Lim , where i1, . . . , in are a finite nonempty family of distinct indices in I. If we send each such x to ϕi1 (xi1 ) ∨ · · · ∨ ϕin (xin ), this is easily seen to have the desired properties. 

There was no analog, in the above result, to the V of Theorem 3, since the variety of semilattices has no proper nontrivial subvarieties. On the other hand, if we wish to get an analog of Theorem 3 with the Li and M again lattices, but for semilattice homomorphisms, rather than isotone maps, we may again start with lattices Li in an arbitrary lattice variety V. For this situation the authors have not been able to prove the full analog of Theorem 3. The difficulty with adapting our proofs of that theorem is that the map ψ of Lemma 2, though isotone, does not respect joins; nor do the variant constructions of Section 3. The map of Lemma 2 does, however, respect joins when the ei are least 0 0 elements in the L¯i. In that case, the composite L → M → M reduces to the map (1) in our sketch of the “easy case” of Theorem 3, and we find that if the ϕi are join-semilattice homomorphisms, that composite will also be one. Hence we get

Proposition 12. Let V be a nontrivial variety of lattices, L = (Li | i ∈ I) a family of lattices in V, L = FreeV L, and ϕi : Li → M a family of join- semilattice homomorphisms from the Li to a common lattice, not necessarily belonging to V. If the image-sets ϕi(Li) have a common lower bound e ∈ M, then there exists a join-semilattice homomorphism ϕ: L → M whose restrictions to the Li ⊆ L are the ϕi. In particular, this is so if M has a least element, or if I is finite and every Li has a least element. 

One could modify this result in the spirit of Theorem 10, assuming that each Li has a retraction %i to a common ideal K on which the ϕi agree. In another direction, the condition in Proposition 12 that there exist a common lower bound e in M to all the ϕi(Li) can be weakened slightly (for I infinite) to say that M contains a chain C such that every ϕi(Li) is bounded below by some member of C. Let us sketch the argument that gets this, by transfinite induction, from the statement as given. First, by passing to a subchain, assume without loss of generality that C is dually well-ordered. Then apply Proposition 12, first, to those Li such that ϕi(Li) is bounded below by the top element, c0, of C, concluding that those ϕi can be factored through 0 some lattice L(0) in V. Then go to the next member, c1, of C, and combine 0 L(0) with all the Li that are bounded below by c1 but not by c0, factoring 0 these together through a lattice L(1) ∈ V; and so on. As in the discussion following (12), we take the union of the preceding steps whenever we hit a limit ordinal. 16 G. M. Bergman and G. Gr¨atzer Algebra univers.

7. Semilattice variants—a harder result

What if we have nothing like the lower-bound condition of Proposition 12? For free products taken in the variety L of all lattices, the analog of that proposition, without the lower bound condition, is obtained in [11, middle of p. 239, “We note finally . . . ”]. Indeed, the map f used in [11] to prove Theorem 3 for L = Free L has the property that f(x ∨ y) = f(x) ∨ f(y) except possibly when x and y are bounded below by elements x(i), y(i) ∈ Li for some i, and ϕi(x(i) ∨ y(i)) > ϕi(x(i)) ∨ ϕi(y(i)). (Cf. [11, p. 238, (ii)].) If the ϕi are join-semilattice homomorphisms, that strict inequality never occurs, so ϕ is also a join-semilattice homomorphism. If V is a nontrivial variety of lattices containing the Li, we do not know whether the corresponding result holds for the free product of the Li in V, but we shall show below that we can get such a result for their free product in the larger class D◦V (definition recalled in (31) and (32) below). Our construction will be similar in broad outline to those used in preceding sections, but the intermediate lattice L0, rather than being a subdirect product, will be a certain lattice of downsets in a direct product. We recall the definition:

If K1 and K2 are classes of lattices, then the class of those lattices L which admit homomorphisms ε: L → L such that L ∈ K , and such 2 2 2 (31) that the inverse image of every element of L2 lies in K1, is denoted K1 ◦ K2.

The class K1 ◦ K2 is often called the product of the classes K1 and K2, but we will not use that name here, to avoid confusion with direct products and free products of lattices. If K1 and K2 are varieties, the class K1 ◦ K2 need not be a variety; but as noted in [13], if K1 and K2 are prevarieties or quasivarieties (classes closed under taking direct products and sublattices; respectively, under taking direct products, ultraproducts, and sublattices), then K1 ◦ K2 will also be a preva- riety, respectively a quasivariety. In particular, if K1 and K2 are varieties, K1 ◦ K2 is, at least, a quasivariety. We also recall the standard notation: The variety of distributive lattices is denoted D. (32)

Now suppose (Li | i ∈ I) is a family of lattices. To begin the construction of the lattice L0 that we shall use in proving our final result, let us adjoin to each Q Li a new top element, 1i, form the direct product (Li + {1i}), and define the subset Q P = {f ∈ (Li + {1i}) | {i | f(i) 6= 1i} is finite but nonempty}. (33)

The condition that {i | f(i) 6= 1i} be nonempty means that we are excluding Q the top element of (Li + {1i}); hence if |I| > 1, P is not a lattice, though it is a lower semilattice. Vol. 00, XX Isotone maps on lattices 17

For each i ∈ I, let us define a map θi : Li → P by

θi(x)(i) = x, θi(x)(j) = 1j for j 6= i. (34) We see that every element of p ∈ P has a representation

p = θi1 (x1) ∧ · · · ∧ θin (xn) with n > 0, (35) unique up to order of terms, where i1, . . . , in are distinct elements of I, and xm ∈ Lim . We now let L0 = the set of all nonempty finitely generated downsets F ⊆ P such (36) that

for all i ∈ I and x, y ∈ Li, if θi(x), θi(y) ∈ F , then θi(x ∨ y) ∈ F . (37) Thus Each element F ∈ L0 is the union of the principal downsets

↓(θi1 (x1) ∧ ... ∧ θin (xn)) determined by its finitely many maximal

elements θi1 (x1) ∧ ... ∧ θin (xn). Moreover, for each i, F can have at (38) most one such maximal element of the form θi(x) (i.e., with n = 1, and with the one meetand arising from Li).

The last sentence above follows from (37): Given distinct θi(x), θi(y) ∈ F , we also have θi(x ∨ y) ∈ F , so θi(x) and θi(y) cannot both be maximal in F . Let us now prove

0 Lemma 13. Let (Li | i ∈ I) be a family of lattices, and let L be constructed as in (33)–(37) above. Then L0, partially ordered by inclusion, is a lattice. (39) For each i ∈ I, the map ξ : L → L0 defined by ξ (x) = ↓ θ (x) is a i i i i (40) lattice homomorphism. 0 If V is any prevariety containing all the Li, then L ∈ D ◦ V. (41) Proof. In verifying (39), the only points that need a moment’s thought are (i) that the intersection F ∩ G of two sets as in (38) remains nonempty and finitely generated; but indeed, in any meet-semilattice, the intersection of two S S nonempty finitely generated downsets ↓ pi and ↓ qj is the nonempty S finitely generated downset ↓(pi ∧ qj); (ii) that the closure operation of (37) cannot produce the element (1i)i∈I ∈/ P ; this follows from the fact that each Li is closed under joins in Li + {1i}; and (iii) that repeated application of that operation when we form a join F ∨ G cannot lead to a violation of finite generation as a downset. This is clear once we observe that in constructing F ∨ G from F ∪ G, it is enough to apply the closure operation of (37) to pairs consisting of one of the finitely many maximal elements of F and one of the finitely many maximal elements of G (and then close as a downset). Statement (40) is easily checked. (Here (37) guarantees that ξi respects joins — that is the point of that condition.) 18 G. M. Bergman and G. Gr¨atzer Algebra univers.

To show (41), let us now adjoin to each Li a bottom element 0i, and define 0 0 maps πi : L → {0i} + Li as follows. For F ∈ L ,

If there are elements x ∈ Li such that θi(x) ∈ F , let πi(F ) be the largest such x (cf. second sentence of (38)). (42)

If there are no such x, let πi(F ) = 0i.

In view of (37), each πi is a homomorphism; hence together they give us a 0 Q homomorphism π : L → ({0i} + Li | i ∈ I) ∈ V. Q We claim that the inverse image under π of each f ∈ ({0i} + Li | i ∈ I) is distributive. Indeed, when we take the join of two elements F,G ∈ π−1(f), we see that for each i, the sets F and G agree in what elements θi(x) they contain, hence there is no occasion for enlarging F ∪ G via (37). So F ∨ G = F ∪ G. We always have F ∧ G = F ∩ G in L0; hence π−1(f) is a lattice of subsets of P under unions and intersections, hence it is distributive. Thus, L0 ∈ D ◦ V, as claimed. 

Now suppose that for each i ∈ I we are given an upper semilattice homo- 0 morphism ϕi : Li → M, for a fixed lattice M. We define ψ : L → M by W ψ(F ) = ( ϕi1 (x1) ∧ ... ∧ ϕin (xn) | θi1 (x1) ∧ ... ∧ θin (xn) ∈ F ). (43) This is formally an infinite join; but it is clearly equivalent to the corresponding join over the finitely many maximal elements of F , hence is defined. We claim that

ψ is a join-semilattice homomorphism. (44)

To see this, note that if we temporarily extend the definition (43) to arbitrary finitely generated downsets F , not necessarily satisfying (37), then we have

ψ(F ∪ G) = ψ(F ) ∨ ψ(G). (45)

Now for F,G ∈ L0, the element F ∨ G is obtained by bringing into F ∪ G elements θi(x ∨ y) where θi(x) ∈ F and θi(y) ∈ G (and the elements they ma- jorize). In this situation, the join defining ψ(F ∪ G) already contains joinands ϕi(x) and ϕi(y), resulting from the presence of θi(x) and θi(y) in F and G, hence its value in M already majorizes ϕi(x) ∨ ϕi(y) = ϕi(x ∨ y). So bringing θi(x ∨ y) into F ∪ G does not increase its image under ψ, establishing (44). Finally, comparing the definition (40) of the ξi and the definition (43) of ψ, we see that

For all i ∈ I, ϕi = ψ ξi. (46)

Now given V as in (41), let L = (Li | i ∈ I) and L = FreeD◦V L. Then the 0 lattice homomorphisms ξi : Li → L are equivalent to a single homomorphism ξ : L → L0; and we see that by taking ϕ = ψ ξ : L → L0 → M, we get our desired result: Vol. 00, XX Isotone maps on lattices 19

Theorem 14. Let L = (Li | i ∈ I) be a family of lattices, and ϕi : Li → M a family of join-semilattice homomorphisms from the Li to a common lattice. Suppose all Li lie in some prevariety V of lattices, and let L = FreeD◦V L. Then there exists a join-semilattice homomorphism ϕ: L → M whose restric- tions to the Li are the ϕi.  Let us show now by example that the lattice L0 constructed in the above proof may fail to lie in V itself. We start with two distributive lattices, namely, the one-element lattice L0 = {e}, and the four-element lattice L1 generated by two elements a and b. Let us use bar notation for the images of these 0 generators under the embeddings ξi : Li → L , so thate ¯ = ξ0(e) = ↓ (e, 11), ¯ 0 a¯ = ξ1(a) = ↓(10, a), b = ξ1(b) = ↓(10, b). We claim that in L , e¯ ∧ (¯a ∨ ¯b) 6= (¯e ∧ a¯) ∨ (¯e ∧ ¯b). (47) Indeed, one finds that the left-hand side of (47) is the principal down-set ↓ (e, a ∨ b), while the right-hand side is (↓ (e, a)) ∪ (↓ (e, b)), a nonprincipal down-set. Hence L0 is not distributive. It is not even modular: one can similarly verify that a copy of N5 is given by the elements (¯e∧a¯)∨(¯e∧¯b)∨(¯a∧¯b), a¯∨(¯e∧¯b), a¯∨(¯e∧(¯a∨¯b)), a¯∨¯b, (¯e∧a¯)∨¯b. (48) Evidence suggesting that the task of extending semilattice homomorphisms from a family of lattices in V to their free product in V is likely to be harder than the corresponding task for isotone maps is [2, Theorem 1] = [12, Theorem 2.8], which says that the injective objects in the category of meet-semilattices are the frames, i.e., the complete lattices satisfying the join-infinite distributive identity. (This result is generalized in [16, Theorem 3.1].) Thus, dually, the injective join-semilattices are the complete lattices satisfying the meet-infinite distributive identity; in particular, they are distributive; so Theorem 14 does not “almost” follow from a general injectivity statement, as Theorem 3 did. (While on the topic of injective objects, what are the injectives in the va- riety of lattices? It is shown in [1, next-to-last paragraph] that the only one is the trivial lattice. This is generalized in [3] to any nontrivial variety V of lattices other than the variety of distributive lattices, and in [14], with a very quick proof, to any class of lattices containing a 3-element chain and a nondistributive lattice.)

8. Questions

The example following Theorem 14 does not mean that there is no way to factor a family of maps as in that theorem through the free product of the Li in V; only that the construction by which we have proved that theorem doesn’t lead to such a factorization. Indeed, for that particular pair of lattices, one does have such a factorization, by the final clause of Proposition 12. So we ask 20 G. M. Bergman and G. Gr¨atzer Algebra univers.

Question 1. For V a general nontrivial variety of lattices, can one prove the full analog of Theorem 3 with join-semilattice homomorphisms in place of isotone maps (i.e., a result like Theorem 14 with V in place of D ◦ V; equivalently, a result like Proposition 12 without the assumptions on lower bounds)? If that result is not true in general, is it true if M also belongs to the given variety V? A counterexample to either version of the above question would probably have to be fairly complicated, in view of Proposition 12. In a different direction, note that in our main result, Theorem 3, the as- sumption that M had a lattice structure did not come into the statement, except to make the concept of isotone map meaningful, for which a struc- ture of order would have sufficed; though the lattice structure was used in the proof. The same observation applies to many of our other results. This suggests a family of questions. Question 2. For each of Lemmas 1 and 2 and Theorems 3, 8, and 10, does the same conclusion hold for a significantly wider class of orders M than the underlying orders of lattices? Likewise, for Proposition 12 and Theorem 14, does the same conclusion hold for a significantly wider class of join-semilattices M than the underlying join-semilattices of lattices? When we showed in Section 3 that our main result could not be proved by a construction with “too much symmetry”, we called on the fact that in a free lattice in the variety L of all lattices, no element is doubly reducible (both a proper meet and a proper join; see sentence following display (13)). Lattices (not necessarily free) with the latter property were considered in [7]. We do not know the answer to Question 3. Are there any nontrivial proper subvarieties V of L such that in every free lattice FreeV(X), no element is doubly reducible? We also record the question arising at the end of Section 2. Question 4. In the first paragraph of Corollary 4, can the isotone map

FreeV L → Free L be taken to be a section (left inverse) to the natural lattice homomorphism Free L → FreeV L?

9. An afterthought

Readers familiar with the study of free products in the variety L of all lattices will be aware of the interplay in that theory between the full forms of lattice terms, and “upper and lower covers” of those terms in the lattices Li Vol. 00, XX Isotone maps on lattices 21

[11, Definition 2], [8, p. 356, Definition 3], [9, Definition 521]. Looking back at the present paper, the authors find that our basic idea can be thought of as that, when constructing isotone maps, one may ignore altogether the full forms of the words, and consider only (something like) their “covers”. This has the advantage of preserving identities satisfied by the Li, and of making the construction fairly transparent. On the other hand, it has required us to introduce the elements ei to serve when “covers” are not available, and also to use the asymmetric definition (5), where for free products in L, one can use formulas that depend (symmetrically) on whether the full form of a reduced lattice word is a meet or a join [11, Definition 2, and p. 238, conditions (ii) and (iii)]. In a general lattice variety V, there is no such meet-join dichotomy for elements of a free product, as shown for D by (13). (Question 3 above essentially asks whether such a dichotomy can hold in any lattice variety other than L.)

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G. M. Bergman 22 G. M. Bergman and G. Gr¨atzer Algebra univers.

University of California, Berkeley, CA, USA e-mail, G. M. Bergman: [email protected] URL, G. M. Bergman: http://math.berkeley.edu/~gbergman/ G. Gratzer¨ Department of Mathematics, University of Manitoba, Winnipeg, MB R3T 2N2, Canada e-mail, G. Gr¨atzer: [email protected] URL, G. Gr¨atzer: http://server.math.umanitoba.ca/homepages/gratzer/